RI_DHS_HCI_TMJC_9649_2023_Prelim_P1_Solutions

FM Prelim Paper 1 (100 marks) 1(a) [2] Let 32f ( ) e 2 xxx    3 2(1) 3 2(2) f (1) (1) e 2 1.1353 (5 s.f.) < 0 f (2) (2) e 2 5.9817 (5 s.f.) > 0           [Alternatively, since 0 < 2e < 1 and 0 < 4e <1, 3 2(1) 2 3 2(2) 4 f (1) (1) e 2 1 e < 0 f (2) (2) e 2 6 e > 0]             Since 22f '( ) 3 +2e 0 xx x x     , f is strictly increasing and continuous. Hence there is exactly one root,  , that lies in [1,2]. (b) [2] Let 32f ( ) e 2 xxx    Using linear interpolation on [1,2] , we have 1 (1)f (2) (2)f (1) 1.1595 1.2 (1 d.p.)f (2) f (1)     (c) [2] 22f '( ) 3 +2e xxx  By Newton-Raphson method, we have 23 1 22 f( ) e 2 f ( ) 3 +2e n n nn n n n n n               Take the initial approximation to be 1 1.2  , 2 1.28058  , 3 1.27609  , 4 1.27607  Checking f (1.2755) 0.00289 < 0 f (1.2765) 0.00215 > 0   That is, 1.276 (3 d.p.)  . 2(a) [2] Sequence converges  as n , nul  and 1nul  8 21 2 ll l   l satisfies the equation 8 21 2 x xx   ,

2 8 21 2 ( 2) 8 21 10 21 0 3 or 7 x xx x x x xx xx          Hence the sequence converges to either 3 or 7. (proven) (b) [4]   1 8 2177 2 8 21 7 2 2 7 0 ( 7 0, 2 0)2 n n n nn n n nn n uu u uu u u uuu                1 7nu  Next, consider 1nnuu  ,   1 8 21 2 8 21 2 2 n n n n n n n n n uu u u u u u u u          2 10 21 2 nn n uu u     37 2 nn n uu u   Since 2nu  , 20nu  . Since 37 nu ,   3 7 0nnuu   . Hence    1 1 37 0 2 0 for 3 7 nn nn n n n n uuuu u u u u            Thus 1nnuu  if 37 nu . Thus, 1 7nnuu  if 37 nu .

(c) [2] From the graph, the sequence strictly increases and converges to 7. Alternatively, Apply (b) repeatedly. 1 1 2 2 23 3 34 3 7 7 37 7 37 7 ... u u u u uu u uu                   Hence we have 1 2 3 4 5 13 ... ... 7 nnu u u u u u u           , and so the sequence strictly increases and converges to 7. 3(a) [3] The characteristic equation for 212 2 0n n nax a x x   is 22 2 1 0aa   , and so for 0a , 2 4 4(2 )(1) 2 ( 2 distinct roots )2(2 ) 2 a a a a a a aa     412 = 1 e22 22 ia a a i a iaa aa       Hence the general solution is 11 22 nn n iix A B aa            .

or 2 ( )cos ( ) sin 442 n n nnx A B A B i a                  (note that A and B are complex numbers) or 2 cos sin , ,442 n n nnx C D C D a                 (b) [3] When 3a , the general solution is 2 cos sin , ,4423 n n nnx C D C D                 0 33xC   1 1 2 3 1 1 3 2 1 3 2 3 2 2 3 x D D D      